hi guys im having trouble figuring this out.

Find the equation of the tangent to the curve

$\displaystyle

y = \frac {x^2} {x-1}

$

when

$\displaystyle

x = -1

$

thank u!

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- Jun 17th 2008, 07:09 AMjvignacioequation of the tangent to the curve
hi guys im having trouble figuring this out.

Find the equation of the tangent to the curve

$\displaystyle

y = \frac {x^2} {x-1}

$

when

$\displaystyle

x = -1

$

thank u! - Jun 17th 2008, 07:20 AMflyingsquirrel
Hello

Let $\displaystyle f(x)=\frac {x^2} {x-1}$

The equation of the tangent to the curve of $\displaystyle f$ at $\displaystyle a$ is $\displaystyle y=f(a)+(x-a)f'(a)$. In this case, $\displaystyle a=-1$ hence what you've to do is simply computing $\displaystyle f(-1)$ and $\displaystyle f'(-1)$.

To find the derivative of $\displaystyle f$, you can use the quotient rule : $\displaystyle \left(\frac{u}{v}\right)'=\frac{u'v-v'u}{v^2}$

Does it help ? - Jun 17th 2008, 07:32 AMjvignacio
yes we did that but what we are confused about is if we use x = -1 for both points?

because we will input -1 into the original formula, then the first point will be (-1, y(original))

then we get the derivative of the original equation then input x into that equation which will give us (-1,y(derivative of original)) ??????

but then if we find m, the denominator will be equal to ZERO which will give us an undefined function.

?? - Jun 17th 2008, 07:38 AMSuumEorum
You find the equation of the tangent by using $\displaystyle y-y_{1}=m(x-x_{1})$ where $\displaystyle x_{1}$ and $\displaystyle y_{1}$ are coordinates for a point on the tangent, in your case, $\displaystyle (-1,f(a))$. m is found by substituting -1 into $\displaystyle y'$, which you get by using the quotient rule.

- Jun 17th 2008, 07:51 AMjvignacio
Yesss i got it! thank u ;)